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\title{\heiti\zihao{2} 习题10.2}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{求下列微分方程的通解 :}
\subsection{$y^{\prime}=\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}$}
\textbf{解}\quad
$$\int \frac{\mathrm{~d} y}{\sqrt{1-y^{2}}}=\int \sqrt{1-x^{2}} \mathrm{~d} x, \arcsin y=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \arcsin x$$

\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=y \sin x$}
\textbf{解}\quad
$$\int \frac{\mathrm{~d} y}{y}=\int \sin x \mathrm{~d} x,y=C e^{-\cos x}$$

\subsection{$3 x y^{\prime}=y $}
\textbf{解}\quad
$$\int \frac{\mathrm{~d} y}{y}=\int \frac{\mathrm{~d} x}{3 x},y=C \sqrt[3]{x}$$

\subsection{$\sqrt{1-y^{2}}=4 x^{3} y y^{\prime}$}
\textbf{解}\quad
$$\int \frac{y \mathrm{~d} y}{\sqrt{1-y^{2}}}=\frac{1}{4} \int \frac{1}{x^{3}} \mathrm{~d} x,-\sqrt{1-y^{2}}=\frac{1}{8 x^{2}}+C$$

\section{求下列齐次方程的通解:}
\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=2 \sqrt{\frac{y}{x}}+\frac{y}{x}$}
\textbf{解}\quad
令$u=\frac{y}{x}$,$xu'+u=y'$,$xu'+u=2\sqrt{u}+u$,有$xu'=2\sqrt{u}$,积分可得$\frac{du}{2\sqrt{u}}=\frac{1}{x}dx$即$\sqrt{u}=\ln x$

即$y=x(\ln x +C)^{2}$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{2 y-x}{2 x-y}$}
\textbf{解}\quad
$y^{\prime}=\frac{2 \frac{y}{x}-1}{2-\frac{y}{x}} \Leftrightarrow x u^{\prime}=\frac{u^{2}-1}{2-u},$ 分离变量可得
$\frac{2-u}{u^{2}-1} \mathrm{~d} u=\frac{1}{x} \mathrm{~d} x, \quad \frac{1}{2} \frac{1}{u-1}-\frac{3}{2} \frac{1}{u+1}=\frac{1}{x} \mathrm{~d} x$
积分可得
$$
	\begin{array}{l}
		\frac{1}{2} \ln |u-1|-\frac{3}{2} \ln |u+1|=\ln x+C_{1} \\
		\ln |u-1|-\ln |u+1|^{3}=\ln x^{2}+2 C_{1}
	\end{array}
$$

即 $\ln \frac{y-x}{(y+x)^{3}}=2 C_{1} \Rightarrow y-x=C(y+x)^{3}$.


\subsection{$\left(1+2 e^{\frac{x}{y}}\right) \mathrm{~d} x+2 e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) \mathrm{~d} y=0$}
\textbf{解}\quad
令 $u=\frac{x}{y}, y u^{\prime}+u=x^{\prime},$ 原题目转换为
$$
	y u^{\prime}=-\frac{u+2 \mathrm{e}^{u}}{1+2 \mathrm{e}^{u}} \Leftrightarrow-\frac{1+2 \mathrm{e}^{u}}{u+2 \mathrm{e}^{u}} \mathrm{~d} u=\frac{1}{u} \mathrm{~d} y
$$

积分可得 $\ln \left|u+2 \mathrm{e}^{u}\right|=-\ln |y|+C_{1},$ 即 $y\left(u+2 \mathrm{e}^{u}\right)=C,$ 所以
$$
	x+2 y \mathrm{e}^{\frac{x}{y}}=C
$$


\subsection{$x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 \sqrt{x y}=y$}
\textbf{解}\quad
令 $u=\frac{y}{x}, x u^{\prime}+u=y^{\prime},$ 当 $x \neq 0$ 时,原方程等价于
$$
	y^{\prime}+2 \sqrt{\frac{y}{x}}=\frac{y}{x} \Leftrightarrow x u^{\prime}=-2 \sqrt{u}
$$

积分可得 $u=(-\ln x+C)^{2},$ 即 $y=x(-\ln x+C)^{2}$

\section{求下列微分方程的通解:}
\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{2 y-x+5}{2 x-y-4}$}
\textbf{解}\quad
$\left|\begin{array}{cc}
		-1 & 2  \\
		2  & -1
	\end{array}\right| \neq 0$,所以方程组 $\left\{\begin{array}{l}2 k-h+5=0 \\ 2 h-k-4=0\end{array}\right.$ 存在唯一解 $\left\{\begin{array}{c}k=-2 \\ h=1\end{array} .\right.$

令$\left\{\begin{array}{l}X=x+1 \\ Y=y-2\end{array}\right.$,$\frac{dY}{dX}=\frac{2Y-X}{2X-Y}=\frac{2 \frac{y}{x}-1}{2-\frac{y}{x}}=\frac{2 u-1}{2-u}$,从而$Xu'=\frac{u^2-1}{2-u}$,从而$\int \frac{2-u}{u^{2}-1} d u=\int \frac{\mathrm{~d} x}{X}$,

解得$\frac{1}{2} \ln |u-1|-\frac{3}{2} \ln |u+1|=\ln X+C$

从而有$Y-X=C(Y+X)^3$.即$y-x-3=C(y+x-1)$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{x-y+1}{x+y-3}$}
\textbf{解}\quad
$\left|\begin{array}{rr}1 & -1 \\ 1 & 1\end{array}\right|\neq 0$,所以方程组 $\left\{\begin{array}{l}h-k+1=0 \\ h-k+3=0\end{array}\right.$ 存在唯一解 $\left\{\begin{array}{l}k=1 \\ h=2\end{array} .\right.$

令$\left\{\begin{array}{l}X=x+1 \\ Y=y+2\end{array},\right.$ 则 $\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{X-Y}{X+Y}$,令$u=\frac{Y}{X}$,由齐次方程的解法可得$\int \frac{1+u}{1-2 u-u^{2}} d u=\int \frac{\mathrm{~d} x}{x}$.

解得:$\ln \left|1-2 u-u^{2}\right|+2\ln |X|=C$,即$(X^2-2YX-Y^2)=C$,即$(x+1)^2-2(y+2)(x+1)-(y+2)^2=C$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3 y-7 x+7}{3 x-7 y-3}$}
\textbf{解}\quad
$\left|\begin{array}{rr}3 & -7 \\ -7 & 3\end{array}\right| \neq 0$ ,所以方程组 $\left\{\begin{array}{l}3 k-7 h+7=0 \\ 3 h-7 k-3=0\end{array}\right.$ 存在唯一解 $\left\{\begin{array}{l}k=0 \\ h=1\end{array} .\right.$

令$\left\{\begin{array}{l}X=x+h=x+1 \\ Y=y+k=y\end{array},\right.$ 则 $\frac{\mathrm{~d} Y}{\mathrm{~d} X}=\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3 \mathrm{Y}-7 \mathrm{X}}{3 \mathrm{X}-7 \mathrm{Y}}=\frac{3 \frac{\mathrm{Y}}{\mathrm{X}}-7}{3-7 \frac{\mathrm{Y}}{\mathrm{X}}},$ 令 $u=\frac{\mathrm{Y}}{\mathrm{X}}, \mathrm{Xu}^{\prime}+u=\mathrm{Y}^{\prime}$
则有
$$
	X u^{\prime}=\frac{7\left(u^{2}-1\right)}{3-7 u} \Rightarrow \int \frac{(3-7 u) d u}{7 u^{2}-1}=\int \frac{\mathrm{~d} x}{X}
$$

从而
$$
	-\frac{3}{2 \sqrt{7}}\left(\ln \left|\frac{\sqrt{7} u+1}{\sqrt{7} u-1}\right|\right)-\frac{1}{2} \ln \left|7 u^{2}-1\right|=\ln X
$$

将$u=\frac{Y}{X}$带回,得$(y-x+1)^2(y+x-1)^5=C$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{x+2 y+1}{2 x+4 y-1}$}
\textbf{解}\quad
令$u=x+2y+1$,得:$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{x+2 y+1}{2 x+4 y-1}=\frac{x+2 y+1}{2(x+2 y+1)-3}=\frac{u}{2 u-3}, \text { 则 } u^{\prime}=1+2 y^{\prime}$

从而$2 \frac{\mathrm{~d} y}{\mathrm{~d} x}=2 y^{\prime}=u^{\prime}-1=\frac{2 u}{2 u-3}$.即$\int \frac{2 u-3}{4 u-3} d u=\int \mathrm{~d} x$,从而$\frac{u}{2}-\frac{3}{8} \ln |4 u-3|-2 x=C$

即$3 \ln |4 x+8 y+1|=8 y-4 x+4+C$

\section{求下列微分方程的通解 :}
\subsection{$y^{\prime} \cos x+y \sin x=1$}
\textbf{解}\quad
$y'+y\tan x=\sec x  $.即$P(x)=\tan x$,$Q(x)=\sec x$.

$$
	\begin{aligned}
	y&=e^{-\int \tan x \mathrm{~d} x}\left(\int \sec x \mathrm{e}^{\int \tan x \mathrm{~d} x} \mathrm{~d} x+C\right)=e^{\ln |\cos x|}\left(\int \sec x e^{\ln \frac{1}{|\cos x|}} \mathrm{~d} x+C\right)\\&=\cos x\left(\int \frac{1}{\cos ^{2} x} \mathrm{~d} x+C\right)=\cos x(\tan x+C)\\&=\sin x+C \cos x
	\end{aligned}
$$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=x \mathrm{e}^{-x}$}
\textbf{解}\quad
$y'+2y=xe^{-x}$,$P(x)=2$,$Q(x)=xe^{-x}$.
$$
\begin{aligned}
y &=\mathrm{e}^{-\int 2 \mathrm{~d} x}\left(\int x \mathrm{e}^{-x} \mathrm{e}^{\int 2\mathrm{~d} x} \mathrm{~d} x+C\right)=\mathrm{e}^{-2 x}\left(\int x \mathrm{e}^{-x} x \mathrm{e}^{2 x} \mathrm{~d} x+C\right) \\
&=\mathrm{e}^{-2 x}\left(\int x \mathrm{e}^{x} \mathrm{~d} x+C\right)=\mathrm{e}^{-2 x}\left[(x-1) \mathrm{e}^{x}+C\right]\\&=(x-1) \mathrm{e}^{-x}+C \mathrm{e}^{-2 x}
\end{aligned}
$$


\subsection{$x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=\sin x$}
\textbf{解}\quad
$y'+\frac{2}{x}y=\frac{\sin x}{x}$,$P(x)=\frac{2}{x}$,$Q(x)=\frac{\sin x}{x}$.
$$
\begin{aligned}
y &=\mathrm{e}^{-\int \frac{2}{x} \mathrm{~d} x}\left(\int \frac{\sin x}{x} \mathrm{e}^{\int \frac{2}{x} \mathrm{~d} x} \mathrm{~d} x+C\right)=\mathrm{e}^{-
\ln x^{2}}
\left(\int \frac{\sin x}{x} \mathrm{e}^{\ln x^{2}} \mathrm{~d} x+C\right) \\
&=\frac{1}{x^{2}}\left(\int x \sin x \mathrm{~d} x+C\right)=\frac{1}{x^{2}}(\sin x-x \cos x+C)=\frac{\sin x}{x^{2}}-\frac{\cos x}{x}+C \frac{1}{x^{2}}
\end{aligned}
$$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}-\frac{1}{1-x^{2}} y=1+x$}
\textbf{解}\quad
$y'+\frac{1}{x^{2}-1}y=1+x$,$P(x)=\frac{1}{x^{2}-1}$,$Q(x)=1+x$
$$
\begin{aligned}
y &=e^{-\int \frac{1}{x^{2}-1} \mathrm{~d} x}\left[\int(1+x) \mathrm{e}^{\int \frac{1}{x^{2}-1} \mathrm{~d} x} \mathrm{~d} x+C\right] \\
&=e^{\frac{1}{2} \ln \left|\frac{x+1}{x-1}\right|}\left[\int(1+x) \mathrm{e}^{\frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|} \mathrm{~d} x+C\right] \\
&=\sqrt{\left|\frac{x+1}{x-1}\right|}\left[\int(1+x) \sqrt{\left|\frac{x-1}{x+1}\right|} \mathrm{~d} x+C\right] \\
&=\sqrt{\frac{x+1}{x-1}}\left(\int \sqrt{x^{2}-1} \mathrm{~d} x+C\right) \\
&=\sqrt{\frac{x+1}{x-1}}\left(\frac{x}{2} \sqrt{x^{2}-1}-\frac{1}{2} \ln \left|x+\sqrt{x^{2}-1}\right|+C\right) \\
&=\frac{x}{2}(x+1)-\frac{1}{2} \sqrt{\frac{x+1}{x-1}} \ln \left|x+\sqrt{x^{2}-1}\right|+C \sqrt{\frac{x+1}{x-1}}
\end{aligned}
$$

\section{求下列微分方程的通解 :}
\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}-\frac{3}{2 x} y=\frac{2 x}{y}$}
\textbf{解}\quad
$yy'-\frac{3}{2x}y^{2}=2x$,令$z=\frac{y^{2}}{2}$,从而$z'-\frac{3}{x}z=2x$.$P(x)=-\frac{3}{x}$,$Q(x)=2x$
$$
\begin{aligned}
z&=e^{\int \frac{3}{x} \mathrm{~d} x}\left[ \int 2xe^{\int -\frac{3}{x} \mathrm{~d} x}+C\right]
\\&=x^{3}[\int \frac{2}{x^{2}}dx +C]\\&=Cx^{3}-2x^2
\end{aligned}
$$
$y^{2}=2Cx^{3}-4x^{2}$.


\subsection{$x \frac{\mathrm{~d} y}{\mathrm{~d} x}+6 y=3 x y^{\frac{4}{3}}$}
\textbf{解}\quad
$y^{-\frac{4}{3}} y^{\prime}+\frac{6}{x} y^{-\frac{1}{3}}=3$,令$z=y^{-\frac{1}{3}}$,则有 $z^{\prime}+\frac{6}{x} z=3$.

如上题,带公式,解得$z=\frac{3}{7}x+\frac{C}{x^{6}}$.从而$y=\left(  \frac{3}{7}x+\frac{C}{x^{6}}   \right)^{-3}$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}+\frac{y}{x}=a(\ln x) y^{2}$}
\textbf{解}\quad
$y^{-2} y^{\prime}+\frac{1}{x} y^{-1}=a \ln x$,令$z=y^{-1}$,则有$z^{\prime}+\frac{1}{x} z=-a \ln x$

如5.1,带公式,解得$z=\frac{1}{2}x\ln x-\frac{1}{4}x+\frac{C}{x}$,从而有$y=\left( \frac{1}{2}x\ln x -\frac{1}{4}x+\frac{C}{x}   \right)^{-1}$.

\section{求下列微分方程的通解 :}
\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\cos (x-y)$}
\textbf{解}\quad
令 $u=x-y \Rightarrow u^{\prime}=1-y^{\prime},$ 所以 $1-u^{\prime}=\cos u,$ 当 $u \neq 2 k \pi$ 时 $,$
$$
\frac{1}{1-\cos u} \mathrm{~d} u=\mathrm{~d} x, \quad \frac{1}{\sin ^{2} \frac{u}{2}} \mathrm{~d}\left(\frac{u}{2}\right)=x+C
$$

积分可得
$$
-\cot \frac{u}{2}=x+C \Leftrightarrow \cot \frac{x-y}{2}=-x-C
$$


\subsection{$x y^{\prime}+y=y(\ln x+\ln y)$;}
\textbf{解}\quad
令 $u=x y \Rightarrow u^{\prime}=y+x y^{\prime},$ 所以 $u^{\prime}=\frac{u}{x} \ln u,$ 当 $u \neq 1$ 时,
$$
\frac{1}{u \ln u} \mathrm{~d} u=\frac{1}{x} \mathrm{~d} x, \quad \ln |\ln u|=\ln |x|+C
$$

积分可得
$$
\ln \left|\frac{\ln u}{x}\right|=C \Rightarrow \frac{\ln x y}{x}=D \Rightarrow \ln x y=\mathrm{~d} x
$$


\subsection{$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{y}{2 x-y^{2}}$}
\textbf{解}\quad
$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{y}{2 x-y^{2}} \Leftrightarrow x^{\prime}=\frac{2 x-y^{2}}{y}=\frac{2 x}{y}-y$ 为一阶线性微分方程,其中 $P=-\frac{2}{y}$
$Q=-y,$ 代人用一阶线性微分方程的通解公式可求得
$$
x=C y^{2}-y^{2} \ln |y|
$$


\subsection{$\cos y \frac{\mathrm \mathrm{~d} y}{\mathrm \mathrm{~d} x}=\sin y+\cos x \sin ^{2} y$}
\textbf{解}\quad
令 $u=\sin y(x) \Rightarrow u^{\prime}=u+\cos x u^{2}$ 为伯努利方程,令 $z=u^{-1},$ 则
$$
z^{\prime}-z=-\cos x
$$

由一阶线性微分方程的通解公式可求得
$$
\begin{array}{c}
z=x\left[\frac{1}{2} \mathrm{e}^{-x}(\cos x-\sin x)+C\right] \\
\text { 故 } \frac{1}{\sin y}=x\left[\frac{1}{2} \mathrm{e}^{-x}(\cos x-\sin x)+C\right] \Leftrightarrow \sin y \cdot x\left[\frac{1}{2} \mathrm{e}^{-x}(\cos x-\sin x)+C\right]=1
\end{array}
$$















\end{document}
\subsection{}
\textbf{解}\quad

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\textbf{证}\quad

\textbf{\textcolor{red}{注}}\quad